most recent 30 from http://mathoverflow.net 2013-06-19T08:19:31Z http://mathoverflow.net/feeds/question/62509 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62509/integral-identity-for-legendre-polynomials Bob Yuncken 2011-04-21T09:06:09Z 2011-04-21T11:05:27Z

<p>How does one prove the following integral identity, where $P_n(x)$ is the $n$th Legendre polynomial? <code>$$ \int_0^1 P_n(2t^2-1) dt = \frac{(-1)^n}{2n+1} $$</code></p> <h3>Notes &amp; Background</h3> <ul> <li><p>A variant of this appears in, for instance, Erdelyi et al "Higher transcendental functions" 10.10(49), but with nothing in the way of explanation.</p></li> <li><p>This comes up in harmonic analysis on $U(3)$, when comparing Gelfand-Tseltin bases associated to different choices of nested sequences $U(3) \supset U(2) \supset U(1)$.</p></li> <li><p>Eventually, I'll be looking for a $q$-analogue, related to harmonic analysis on $U_q(3)$, so a proof that will transport well would be my true desire.</p></li> </ul>

http://mathoverflow.net/questions/62509/integral-identity-for-legendre-polynomials/62520#62520 Bob Yuncken 2011-04-21T11:05:27Z 2011-04-21T11:05:27Z

<p>Nice idea. As far as I'm concerned, the above comments are "answers", since they check out. I might as well record the details: <code>$$ \int_0^1 \frac{dx}{\sqrt{1-2(2x^2-1)t + t^2}} = \frac{1}{2\sqrt{t}} \int_0^1 \frac{dx}{\sqrt{\frac{(1+t)^2}{4t} -x^2}} = \frac{1}{2\sqrt{t}}\arcsin\left(\frac{\scriptstyle 2\sqrt{t}}{\scriptstyle1+t}\right). $$</code> The half-angle formula for $\sin$ reduces this to $$ \frac{1}{\sqrt{t}} \arcsin \sqrt{\frac{t}{1+t}} = \frac{1}{\sqrt{t}} \arctan \sqrt{t}, $$ which has the desired power series expansion.</p> <p>Thanks.</p>